Question

How do you solve by completing the square: `x^2 + 8x + 2 = 0`?

Answer 1

The answer is `x = -4` `+-` `sqrt (14)`

The general form of a trinomial is `ax^2+bx+c=0"` The letter `c` is the constant.

Solve the trinomial `x^2+8x+2=0`

First move the constant to the right side by subtracting 2 from both sides.

`x^2+8x=-2`

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

`((8)/(2))^2=(4)^2=16`

`x^2+8x+16 = -2+16`

`x^2+8x+16=14`

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

`(x+4)^2=14`

Take the square root of each side and solve.

`x+4`=`+-` `sqrt (14)`

`x = -4` `+-` `sqrt (14)`

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

Answer 2

The answer is `x = -4` `+-` `sqrt (14)`

The general form of a trinomial is `ax^2+bx+c=0"` The letter `c` is the constant.

Solve the trinomial `x^2+8x+2=0`

First move the constant to the right side by subtracting 2 from both sides.

`x^2+8x=-2`

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

`((8)/(2))^2=(4)^2=16`

`x^2+8x+16 = -2+16`

`x^2+8x+16=14`

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

`(x+4)^2=14`

Take the square root of each side and solve.

`x+4`=`+-` `sqrt (14)`

`x = -4` `+-` `sqrt (14)`

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm